Question

A nucleus of mass number 220 initially at rest emits an alpha particle. If the $Q$ value of the reaction is $5.5MeV$, then the kinetic energy of the alpha particle is

• A. $6.5MeV$
• B. $5.5MeV$
• C. $5.4MeV$
• D. $2.0MeV$

Answer 1

The correct option is C $5.4MeV$

Let $K{E}_{\alpha }$ be the kinetic energy of the alpha particle and $K{E}_d$ be the kinetic energy of daughter nuclei.
Since $Q$ value of the reaction is $5.5MeV$,
$K{E}_{\alpha }+K{E}_d=5.5MeV$ --------(1)

From momentum conservation,
initial momentum = final momentum
$0=\overrightarrow{p_{\alpha }}+\overrightarrow{p_d}$

Hence, $p_{\alpha }^2=p_d^2$
or $2\times m_{\alpha }\times K{E}_{\alpha }=2\times m_d\times K{E}_d$
($\because p=\sqrt{2mKE}$)

Since it's an alpha decay, $m_d=220-4=216$
and $m_{\alpha }=4$

Putting these values, we get
$\frac{K{E}_{\alpha }}{K{E}_d}=\frac{m_d}{m_{\alpha }}=54$ -----(2)

From equations (1) and (2),
$K{E}_{\alpha }+\frac{K{E}_{\alpha }}{54}=5.5MeV$
$\Rightarrow K{E}_{\alpha }=5.4MeV$